| Topic: |
Science > Physics |
| User: |
"kevin" |
| Date: |
11 Sep 2006 05:08:00 PM |
| Object: |
help with problem |
this might be a little elemtary, but it has become a confusing problem
for me. If a wieght lifter lifts wieght and then immediately drops the
wieght, the total amount of work done (by person onto weight) is simply:
W = mgh * (# reps)
But what if the weight lifter slowly, at constant velocity, lowers the
wieght back onto the floor after lifting it? The Work on the way up is
the same as before, but on the way down the displacement is in the
opposite direction of this, so does this mean the net work is zero? The
work that something does is path dependent (i think), so it seems that
the person should have performed a net amount of work. However, the
weight itself is returned to its same level of potential energy. So the
wieght itself is in the same energetic state as before, but the person
has expended energy during the process.
The question is, what is the total amount of work done by the person
onto the weight during this process? Mathematically, solving it seems to
say 0, but this seems very counter intuitive to me.
Thanks for any cosiderations
kevin
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| User: "CWatters" |
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| Title: Re: help with problem |
12 Sep 2006 08:21:06 AM |
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"kevin" <kbbb@mail.utexas.edu> wrote in message
news:ee4mnv$ip0$1@geraldo.cc.utexas.edu...
this might be a little elemtary, but it has become a confusing problem
for me. If a wieght lifter lifts wieght and then immediately drops the
wieght, the total amount of work done (by person onto weight) is simply:
W = mgh * (# reps)
But what if the weight lifter slowly, at constant velocity, lowers the
wieght back onto the floor after lifting it? The Work on the way up is
the same as before, but on the way down the displacement is in the
opposite direction of this, so does this mean the net work is zero?
The net work done on the weight is indeed zero no matter how it arrives back
at it's rest position. The net work done by the weight lifter is another
matter. He expends energy on the way up but can't recover that energy on the
way down. His body is like an electric car without regenerative braking. You
can accelerate upto speed but you have to waste that energy heating up the
brakes when you stop.
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| User: "Sorcerer" |
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| Title: Re: help with problem |
11 Sep 2006 08:47:53 PM |
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"kevin" <kbbb@mail.utexas.edu> wrote in message
news:ee4mnv$ip0$1@geraldo.cc.utexas.edu...
| this might be a little elemtary, but it has become a confusing problem
| for me. If a wieght lifter lifts wieght and then immediately drops the
| wieght, the total amount of work done (by person onto weight) is simply:
|
| W = mgh * (# reps)
|
| But what if the weight lifter slowly, at constant velocity, lowers the
| wieght back onto the floor after lifting it? The Work on the way up is
| the same as before, but on the way down the displacement is in the
| opposite direction of this, so does this mean the net work is zero?
Yes. Nothing has been done to the weight, it is back where it was.
Even if the weightlifter holds the weight above his head, sweating,
he is doing no further work on the weight. His muscles will tire,
but the work the weight can do in bending the floorboards is the
same.
The
| work that something does is path dependent (i think), so it seems that
| the person should have performed a net amount of work. However, the
| weight itself is returned to its same level of potential energy. So the
| wieght itself is in the same energetic state as before, but the person
| has expended energy during the process.
| The question is, what is the total amount of work done by the person
| onto the weight during this process?
Nothing more than the potential energy he gave it. He can rest it on
a shelf and walk away or he can stand and sweat, it makes no
difference to the weight.
Mathematically, solving it seems to
| say 0, but this seems very counter intuitive to me.
|
| Thanks for any cosiderations
| kevin
Trust mathematics, intuition is fallible. When you do you'll
gain new intuitions.
Androcles
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| User: "Timo Nieminen" |
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| Title: Re: help with problem |
11 Sep 2006 05:58:43 PM |
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|
On Mon, 11 Sep 2006, kevin wrote:
this might be a little elemtary, but it has become a confusing problem
for me. If a wieght lifter lifts wieght and then immediately drops the
wieght, the total amount of work done (by person onto weight) is simply:
W = mgh * (# reps)
But what if the weight lifter slowly, at constant velocity, lowers the
wieght back onto the floor after lifting it? The Work on the way up is
the same as before, but on the way down the displacement is in the
opposite direction of this, so does this mean the net work is zero? The
work that something does is path dependent (i think), so it seems that
the person should have performed a net amount of work. However, the
weight itself is returned to its same level of potential energy. So the
wieght itself is in the same energetic state as before, but the person
has expended energy during the process.
The question is, what is the total amount of work done by the person
onto the weight during this process? Mathematically, solving it seems to
say 0, but this seems very counter intuitive to me.
OK, the weight ends up on the floor with the same potential energy that it
started with, and the same kinetic energy (ie zero) that it started with,
so the total work done on the weight is zero.
W=mgh*reps is not the total work done on the weight; it's the total work
done in the lifting part of the motion. If the weightlifter drops the
weight each time instead of slowly lowering it, then it's the total work
done by the weightlifter.
If the weightlifter drops the weight, then the
floor does (negative) work on the weigh when it hits the floor, reducing
its PE+KE back to the original value. If the weightlifter slowly lowers
the weight, the weightlifter does that same amount of negative work (over
a longer distance than the floor does), and the total work done by the
weightlifter on the weight is zero.
Note that as the weightlifter lowers the weight, the work done on the
weight is negative. Yet the weightlifter has to use a lot of energy to
lower the weight. Why? Individual muscle cells contract repeatedly,
pulling in the direction of contraction, doing work. They then relax and
return to their original length, without getting any of that energy back.
This makes muscles very inefficient, and you can't get energy back from
them by doing negative work - you just end up using even more energy. You
could get some, or even most, of the energy used to lift the weight beack
if you used some kind of mechanical system, such as a spring, a weight on
a pulley, a hydraulic lift etc. Muscles, no.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
.
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| User: "kevin" |
|
| Title: Re: help with problem |
11 Sep 2006 06:29:36 PM |
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|
OK, to clarify...
Work depends on vectors, and since the dispacement is negative on the
way down, the total addition of work is basically 0. The chemical PE
of the person is the source of lifting energy, and must also
be used on the way down (unlike for a mechanical device) to slow down
the falling of the weight. So although the releasing of the weights
PE brings the work down to 0, the energy used continues to grow.
This is what confused me. The energy used by the person must then not be
turned into WORK, but in the end, heat energy.
so for the person:
using : q + W = dU
-(heat given off) + 0 = -(change in internal energy of person) = amount
of energy used to lift and lower weight
for wieght :
W = [mg h + mg (-h)] * reps = 0
assume q = 0
0 + 0 = 0
thanks
kevin
Timo Nieminen wrote:
On Mon, 11 Sep 2006, kevin wrote:
this might be a little elemtary, but it has become a confusing problem
for me. If a wieght lifter lifts wieght and then immediately drops the
wieght, the total amount of work done (by person onto weight) is simply:
W = mgh * (# reps)
But what if the weight lifter slowly, at constant velocity, lowers the
wieght back onto the floor after lifting it? The Work on the way up is
the same as before, but on the way down the displacement is in the
opposite direction of this, so does this mean the net work is zero? The
work that something does is path dependent (i think), so it seems that
the person should have performed a net amount of work. However, the
weight itself is returned to its same level of potential energy. So the
wieght itself is in the same energetic state as before, but the person
has expended energy during the process.
The question is, what is the total amount of work done by the person
onto the weight during this process? Mathematically, solving it seems to
say 0, but this seems very counter intuitive to me.
OK, the weight ends up on the floor with the same potential energy that it
started with, and the same kinetic energy (ie zero) that it started with,
so the total work done on the weight is zero.
W=mgh*reps is not the total work done on the weight; it's the total work
done in the lifting part of the motion. If the weightlifter drops the
weight each time instead of slowly lowering it, then it's the total work
done by the weightlifter.
If the weightlifter drops the weight, then the
floor does (negative) work on the weigh when it hits the floor, reducing
its PE+KE back to the original value. If the weightlifter slowly lowers
the weight, the weightlifter does that same amount of negative work (over
a longer distance than the floor does), and the total work done by the
weightlifter on the weight is zero.
Note that as the weightlifter lowers the weight, the work done on the
weight is negative. Yet the weightlifter has to use a lot of energy to
lower the weight. Why? Individual muscle cells contract repeatedly,
pulling in the direction of contraction, doing work. They then relax and
return to their original length, without getting any of that energy back.
This makes muscles very inefficient, and you can't get energy back from
them by doing negative work - you just end up using even more energy. You
could get some, or even most, of the energy used to lift the weight beack
if you used some kind of mechanical system, such as a spring, a weight on
a pulley, a hydraulic lift etc. Muscles, no.
.
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|
| User: "Timo Nieminen" |
|
| Title: Re: help with problem |
11 Sep 2006 08:05:12 PM |
|
|
On Mon, 11 Sep 2006, kevin wrote:
OK, to clarify...
Work depends on vectors, and since the dispacement is negative on the
way down, the total addition of work is basically 0. The chemical PE
of the person is the source of lifting energy, and must also
be used on the way down (unlike for a mechanical device) to slow down
the falling of the weight. So although the releasing of the weights
PE brings the work down to 0, the energy used continues to grow.
This is what confused me. The energy used by the person must then not be
turned into WORK, but in the end, heat energy.
Yes.
This point should be explicitly discussed in class. Students _know_ that
they have to keep using energy to hold a weight stationary in the air.
Their lecturer tells them "no work is done on the weight". Oops, this
doesn't seem to match! Better to explain and not leave students confused.
Likewise, why we never actually see Newton's 1st law in action in everyday
life should be (briefly) discussed.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
.
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| User: "Sue..." |
|
| Title: Re: help with problem |
11 Sep 2006 05:42:02 PM |
|
|
kevin wrote:
this might be a little elemtary, but it has become a confusing problem
for me. If a wieght lifter lifts wieght and then immediately drops the
wieght, the total amount of work done (by person onto weight) is simply:
W = mgh * (# reps)
But what if the weight lifter slowly, at constant velocity, lowers the
wieght back onto the floor after lifting it? The Work on the way up is
the same as before, but on the way down the displacement is in the
opposite direction of this, so does this mean the net work is zero? The
work that something does is path dependent (i think), so it seems that
the person should have performed a net amount of work. However, the
weight itself is returned to its same level of potential energy. So the
wieght itself is in the same energetic state as before, but the person
has expended energy during the process.
The question is, what is the total amount of work done by the person
onto the weight during this process? Mathematically, solving it seems to
say 0, but this seems very counter intuitive to me.
Very close to zero is correct.
Consider the weight lifter in microgravity on the ISS.
He works only at accelerating and decelerating
the weights, and this only because his muscles
and tendons are not perfect springs.
On the earths surface the configuration of
the springs and weights is a bit different
but still, he is just accelerting a weight.
A pendulm clock would run forever if you
with no energy could reduce all the losses.
We don't consider the heating of a clock
beaing work and neither do we consider
heating of the weight lifters muscles,
http://hyperphysics.phy-astr.gsu.edu/hbase/pend.html
Sue...
Thanks for any cosiderations
kevin
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